Zener Diodes. Specifying and modeling the zener diode. - Diodes operating in the breakdown region can be used in the design of voltage regulators.
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1 Zener Diodes - Diodes operating in the breakdown region can be used in the design of voltage regulators. Specifying and modeling the zener diode Dynamic resistance, r Z a few ohms to a few tens of ohms V ri z Zener breakdown voltage, V Z a few volts to a few hundreds of volts Figure 3.16 Circuit symbol for a zener diode. In Fig. 3.17, the voltage of Q-point VZ VZ 0 ri z Z VZ VZ 0 ri z Z Figure 3.18 Model for the zener diode. Figure 3.17 The diode i v characteristic with the breakdown region shown in some detail. 22
2 Zener Diodes Use of the zener as a shunt regulator The change in V O corresponding to a 1-V change in V S [ mv /V] : The change in V O corresponding to a 1-mA change in I L [ mv / ma ] : Assuming R, r Z R L, R rz V O VZ 0 VS IL( R // rz ) R r R r Thus, Line regulation Z rz R r z Z V Line regulation V V Load regulation I O S O L Load regulation ( r // R) z 23
3 Zener Diodes Ex 3.7 V Z =6.8V at I Z =5mA, r Z =20Ω, and I ZK =0.2mA. V V ri V Z 0 Z z Z a) With no load and with V + =10V, the current through the zener is given by + V VZ IZ I 6.35 ma, Rrz V V Ir V O Z 0 Z z b) The change in V O resulting from the ±1V change in V V rz O V mv Rr Line regulation z V O V 38.5 mv/v c) The change in V O resulting from connecting a load resistance R L that draws a current I L =1mA V ri mv O z Z Load regulation V O I L 20 mv/ma Figure 3.19 (a) Circuit for Example 3.8. (b) The circuit with the zener diode replaced with its equivalent circuit model. 24
4 Zener Diodes d) The change in V O when R L =2 kω I L 6.8V/ 2k 3.4mA, I 3.4 ma Z V ri mv O z Z e) When R L =0.5kΩ, I L = 6.8/0.5=13.6mA > I. This is impossible, therefore V O 6.8V, and the zener must be cut off V RL O V 10 5 V RR L < 6.8V f) The minimum value of R L for which the diode still operates in the breakdown region. I I 0.2mA, V V V 6.7V Z ZK Z ZK Z Imin I L 6.7 RL,min 1.5 k ma, = =4.4 ma * Recently, zener diodes have been replaced in IC voltage-regulators. 25
5 Rectifier Circuits The half-wave rectifier (diode off) In many applications, V D = 0.7V or 0.8V (diode on) Figure 3.20 Block diagram of a dc power supply. Peak inverse voltage the largest reverse voltage that is expected to appear across the diode. PIV = V s to select a diode that has a reverse breakdown voltage at least 50% greater than the expected PIV Figure 3.21 (a) Half-wave rectifier. (b) Transfer characteristic of the rectifier circuit. (c) Input and output waveforms, assuming that r D R. 26
6 Rectifier Circuits The full-wave rectifier The primary voltage > 0 D 1 on, D 2 off The primary voltage < 0 D 1 off, D 2 on During the +tive half-cycle, the voltage at the cathode of D 2 is v O, and that at its anode is -v S. Thus the reverse voltage across D 2 will be (v O +v S ), which will reach its maximum when v O is at its peak value of (V s - V D0 ), and v S is at at its peak value of V s. PIV = 2V s - V D0 Figure 3.22 Full-wave rectifier utilizing a transformer with a center-tapped secondary winding: (a) circuit; (b) transfer characteristic assuming a constant-voltage-drop model for the diodes; (c) input and output waveforms. 27
7 Rectifier Circuits The bridge rectifier v S > 0 D1, D2 on D3, D4 off v S < 0 D1, D2 off D3, D4 on When v S > 0, = + = - D3, r O D2, f S D1, f Thus PIV = Vs 2VD VD Vs VD Figure 3.23 The bridge rectifier: (a) circuit; (b) input and output waveforms. The rectifier with a filter capacitor Input voltage v I =V p sin wt Assuming that the diode is ideal, t < T/4 diode on, v O = v I (charging capacitor) t T/4 diode off, v O = V p Figure 3.24 (a) A simple circuit used to illustrate the effect of a filter capacitor. (b) Input and output waveforms assuming an ideal diode. Note that the circuit provides a dc voltage equal to the peak of the input sine wave. The circuit is therefore known as a peak rectifier or a peak detector. 28
8 Rectifier Circuits The peak rectifier Assuming that the diode is ideal, t < 0 diode on, v O = v I (capacitor charging) 0 < t < T -t diode off, v O = V p e -t/rc (capacitor discharging) T -t < t < T diode on, v O = v I (capacitor charging) Under the assumption that RC T, Load current Diode current i L / R / V V V T CR p r p e O d i i i C i dt I D C L L Output dc voltage, V o = V p -V r /2, where V r is ripple voltage Output dc current, I L = V o /R V p /R During the diode-off interval, v o = V p e -t/cr If Δt T, at the end of the discharge interval, Since RC T, e -t/cr 1-T/CR T Vp IL Vr Vp CR fcr fc 0 T Figure 3.25 Voltage and current waveforms in the peak rectifier circuit with CR T. 29
9 Rectifier Circuits Conduction interval t where w = 2p f = 2p / T. Since wt is small, Charge that the diode supplies to the capacitor where, Charge that the capacitor loses To equate two charges, Peak value of the diode current 0 T 30
10 Rectifier Circuits Vp 100 C 83.3 F 3 V fr r The diode conducts for of the cycle. In case of the full-wave rectifier Vp Vr 2 fcr Figure 3.26 Waveforms in the full-wave peak rectifier. 31
11 Limiting and Clamping Circuits Limiting circuits v I L - /K, v o = L - L - /K v I < L + /K, v o = K v I L + /K v I, v o = L + Figure 3.28 General transfer characteristic for a limiter circuit. Figure 3.29 Applying a sine wave to a limiter can result in clipping off its two peaks. Figure 3.31 A variety of basic limiting circuits. 32
12 Limiting and Clamping Circuits The clamped capacitor Capacitor voltage v I < 0 diode on v C = -v I (to charge capacitor) v I 0 diode off v C = minus peak of v I Output voltage v O = v I + v C * Reversing the diode polarity will provide an output waveform whose highest peak is clamped to 0V. Figure 3.32 The clamped capacitor or dc restorer with a square-wave input and no load. Figure 3.33 The clamped capacitor with a load resistance R. 33
13 Limiting and Clamping Circuits The voltage doubler C 1 and D 1 - clamping circuit D 2 and C 2 - peak detector Figure 3.34 Voltage doubler: (a) circuit; (b) waveform of the voltage across D 1. 34
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